HDU - 1102 Constructing Roads (最小生成树 prim)

墨蓝 2022-05-18 05:25 135阅读 0赞

#                                       Constructing Roads #

### Problem Description ###

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.  
  
We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

### Input ###

The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within \[1, 1000\]) between village i and village j.  
Then there is an integer Q (0 <= Q <= N \* (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

### Output ###

You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

### Sample Input ###

0 990 692

990 0 179

692 179 0

1 2

### Sample Output ###

179

**题意描述：**  
给出了各个村庄间的距离，并且有的村庄已经联通，求使各个村庄能互相连通最小长度；最小生成树模板题。

**解题思路：**  
存入各村庄的距离，将村庄相通的距离更新为0，直接运用prim算法模板；

#include<stdio.h>
    #include<string.h>
    # define inf 99999999
    int map[1010][1010];
    int book[1010],dis[1010];
    int main()
    {
    	int n,m,i,j,u,v,min,a,b,sum;
    	while(scanf("%d",&n)!=EOF)
    	{
    		for(i=1;i<=n;i++)
    			for(j=1;j<=n;j++)
    				scanf("%d",&map[i][j]);
    		scanf("%d",&m);
    		for(i=1;i<=m;i++)
    		{
    			scanf("%d%d",&a,&b);//已连通的村庄距离更新为0 
    			map[a][b]=0;
    			map[b][a]=0;
    		}
    		memset(book,0,sizeof(book));
    		for(i=1;i<=n;i++)
    			dis[i]=map[1][i];
    		book[1]=1;
    		sum=0;
    		for(i=1;i<n;i++)//prim算法 
    		{
    			min=inf;
    			for(j=1;j<=n;j++)
    			{
    				if(book[j]==0&&dis[j]<min)
    				{
    					min=dis[j];
    					u=j;
    				}
    			}
    			book[u]=1;
    			sum=sum+dis[u];
    			for(v=1;v<=n;v++)
    			{
    				if(book[v]==0&&dis[v]>map[u][v])
    					dis[v]=map[u][v];
    			}
    		}
    		printf("%d\n",sum);
    	}
    	return 0;
    }